\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc} % 添加这一行来引入 calc 库
\usepackage{ctex}

\begin{document}
Proof Idea

We will use a geometric approach to prove the trigonometric identity $\sin(\alpha + \beta) = \sin\alpha\cos\beta+\cos\alpha\sin\beta$. We'll construct a geometric figure that contains angles $\alpha$, $\beta$, and $\alpha + \beta$, and then derive the identity through the relationships between line - segment lengths.

Specific Proof Process

1. **Construct the geometric figure**:

    - Draw a triangle $\triangle ABC$ where OC = 1 and $OC\perp AB$ 
    \begin{center}
    \begin{tikzpicture}[scale = 2]
      % Define points
      \coordinate (O) at (0,0);
      \coordinate (C) at (0,3);
      \coordinate (A) at (-3,0);
      \coordinate (B) at (1.7,0);

      % Draw triangles and line segments
      \draw (A) -- (C) -- (B) -- cycle;
      \draw (O) -- (C);

      % Label points
      \node[below left] at (A) {$A$};
      \node[below right] at (B) {$B$};
      \node[above] at (C) {$C$};
      \node[below] at (O) {$O$};

      % Label angles
      \draw[red] (-0.2,2.8) arc (50:110:-0.2);
      \node[red] at (-0.18,2.6) {$\alpha$};
      \draw[blue] (0.3,2.5) arc (120:60:-0.3);
      \node[blue] at (0.15,2.2) {$\beta$};
    \end{tikzpicture}
    \end{center}

    I found 
    
    $AO = \tan\alpha$
    
    $BO = \tan\alpha$

    and
    
    $AB = AO + BO$
    
    $= \tan\alpha + \tan\beta$ 
                  
                  $= \frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta} $
     
                  $=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}$

    Here I found,

    $\sin\alpha\cos\beta+\cos\alpha\sin\beta$

    is just what I want.

    But How to get the solution?
    

2. **Calculate $\sin(\alpha + \beta)$**:

    - According to the definition of the sine function, in $\triangle ABC$,

    
    Draw a line segment inside $\triangle ABC$ where $AD\perp BC$ 
    
    $\sin(\alpha + \beta)=\frac{AD}{AC}$.

     

    

\begin{center}
    \begin{tikzpicture}[scale = 2]
      % Define points
      \coordinate (O) at (0,0);
      \coordinate (C) at (0,3);
      \coordinate (A) at (-3,0);
      \coordinate (B) at (1.7,0);
      % 计算 D 点坐标，D 是 A 到 BC 的垂足
      \coordinate (D) at ($(B)!(A)!(C)$); 

      % Draw triangles and line segments
      \draw (A) -- (C) -- (B) -- cycle;
      \draw (O) -- (C);
      \draw (A) -- (D); % 绘制 AD

      % Label points
      \node[below left] at (A) {$A$};
      \node[below right] at (B) {$B$};
      \node[above] at (C) {$C$};
      \node[below] at (O) {$O$};
      \node[below] at (D) {$D$}; % 标注 D 点

      % Label angles
      \draw[red] (-0.2,2.8) arc (50:110:-0.2);
      \node[red] at (-0.18,2.6) {$\alpha$};
      \draw[blue] (0.3,2.5) arc (120:60:-0.3);
      \node[blue] at (0.15,2.2) {$\beta$};
      % 标记直角
      % 标记旋转 45 度的直角
      \begin{scope}[rotate around={120:(D)}]
        \draw ($(D)+(0.1,0)$) -- ($(D)+(0.1,0.1)$) -- ($(D)+(0,0.1)$); 
      \end{scope}
    \end{tikzpicture}
    \end{center}

 For AC,

    $AC = \frac{CO}{\cos\alpha} = \frac{1}{\cos\alpha}$

    since CO = 1

    But AD = ?

    The area of $\triangle ABC$ can be calculated in 2 method,

    $area = \frac{AB*CO}{2} = \frac{AB}{2}$

    or

    $area = \frac{AD*BC}{2}$

    so

    $AD = \frac{AB*CO}{BC}$

    so

        
    $\sin(\alpha + \beta)=\frac{AD}{AC}$.

    $=\frac{AB*CO}{BC*AC}$

    For AC,

    $AC = \frac{CO}{\cos\alpha} = \frac{1}{\cos\alpha}$

    For BC,

    $BC = \frac{CO}{\cos\beta} = \frac{1}{\cos\beta}$

        For AB,

    $AB=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}$

        For CO,

    CO = 1

  then,

   $\sin(\alpha + \beta)=\frac{\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{1}{\cos\beta}*\frac{1}{\cos\alpha}}=\sin\alpha\cos\beta+\cos\alpha\sin\beta$.



\end{document}